Results of testing for the presence of pollutants in a local stream have a mean of 10 mg/L and a standard deviation of 2 mg/L. Six samples of water collected from the stream result in the following...

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There are two ways to solve this.
Assuming population standard deviation is same
as before
Given Data:
12.7, 15.1, 9.5, 13.7, 19.6, 16.4
We calculate sample mean Sample Mean, x̄ =
∑
(x)
n
= 876 = 14.5
Test: µ > 10 with normal distn
Hypothesis test:
H0 : µ ≤ 10 (Null Hypothesis, H0 : µ = 10 is also correct)
Ha : µ > 10 (Alternative Hypothesis, also called H1)
This is upper tailed test (one tailed test).
x = 14.5
σ = 2
n = 6
Significance Level, α = 0.05 (If no value is given, we take level of 0.05)
Test statistic z? = x− µ
σ/
√
n
Test Statistic, z? = 14.5− 10
2/
√
6
≈ 5.511352 ≈ 5.51
Test Statistic, z? = 5.51
P-value is in direction of Alternative hypothesis.
Since Ha : µ > 10, P-value= P (Z > 5.51)
P-value is Area to the right of test statistic = 5.51
P − value = 0.0000
test statistic = 5.51
0
P-value = P (Z > 5.51) = 0.0000(from z-table)
P-value = 0.0000
Decision Rule (rejection criteria): Reject H0 if p-value < α
Decision: Since 0.0000 < 0.05, we reject the null hypothesis.
Using excel function...